18  Ray Optics

18.1 Syllabus inquiry question

  • How can light be modelled to predict image formation?
Feynman Insight

From The Feynman Lectures on Physics, Vol I, Chapter 33:

The ray model works because light travels in straight lines in uniform media, and refraction changes those lines in a precise, predictable way.

18.2 Learning Objectives

  • Apply laws of reflection and refraction.
  • Use Snell’s law to predict angles.
  • Apply thin lens equation and magnification.
  • Distinguish real and virtual images.

18.3 Content

18.3.1 The Ray Model of Light

In ray optics, we model light as straight lines (rays) that travel in uniform media and change direction at boundaries.

Key assumptions: - Light travels in straight lines - Rays are perpendicular to wavefronts - Valid when obstacles are much larger than wavelength

18.3.2 Reflection

The law of reflection states:

\[\theta_i = \theta_r\]

  • Angle of incidence (\(\theta_i\)): Angle between incident ray and normal
  • Angle of reflection (\(\theta_r\)): Angle between reflected ray and normal
  • The normal is perpendicular to the surface

18.3.3 Interactive: Reflection from a Plane Mirror

18.3.4 Refraction and Snell’s Law

Refraction occurs when light crosses a boundary between two media with different optical densities.

Snell’s Law: \[n_1 \sin\theta_1 = n_2 \sin\theta_2\]

where \(n\) is the refractive index of each medium.

Refractive Index

The refractive index measures how much light slows down in a medium: \[n = \frac{c}{v}\] where \(c\) is the speed of light in vacuum and \(v\) is the speed in the medium.

18.3.5 Common Refractive Indices

Material Refractive Index (\(n\))
Vacuum 1.00 (exactly)
Air 1.00 (approximately)
Water 1.33
Glass 1.50 (typical)
Diamond 2.42

18.3.6 Interactive: Refraction at a Boundary

Light bending as it enters a denser medium:

Key observation: Light bends toward the normal when entering a denser (higher \(n\)) medium.

18.3.7 Total Internal Reflection

When light travels from a denser to a less dense medium, there exists a critical angle \(\theta_c\) beyond which all light is reflected internally.

\[\sin\theta_c = \frac{n_2}{n_1}\]

Applications

Total internal reflection is used in: - Optical fibres (telecommunications) - Prisms in binoculars - Endoscopes (medical imaging)

18.3.8 Thin Lens Equation

For thin lenses, the object distance (\(u\)), image distance (\(v\)), and focal length (\(f\)) are related by:

\[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\]

Sign conventions:

  • Real objects: \(u > 0\) (in front of lens)
  • Real images: \(v > 0\) (on opposite side from object)
  • Virtual images: \(v < 0\) (same side as object)
  • Converging lens: \(f > 0\)
  • Diverging lens: \(f < 0\)

18.3.9 Magnification

Magnification describes how the image size compares to the object size:

\[m = \frac{h_i}{h_o} = -\frac{v}{u}\]

  • \(|m| > 1\): Image larger than object
  • \(|m| < 1\): Image smaller than object
  • \(m > 0\): Upright image
  • \(m < 0\): Inverted image

18.3.10 Interactive: Converging Lens Ray Diagram

18.3.11 Real vs Virtual Images

Property Real Image Virtual Image
Can be projected Yes No
Light rays Actually converge Appear to diverge from a point
Orientation Inverted Upright
Sign of \(v\) Positive Negative

18.4 Worked Examples

18.4.1 Example 1: Snell’s Law

Light travels from air (\(n = 1.00\)) into glass (\(n = 1.50\)) at an incident angle of 30°.

Solution:

  1. Apply Snell’s Law: \(n_1 \sin\theta_1 = n_2 \sin\theta_2\)

  2. Rearrange: \(\sin\theta_2 = \frac{n_1}{n_2} \sin\theta_1 = \frac{1.00}{1.50} \times \sin30°\)

  3. Calculate: \(\sin\theta_2 = 0.667 \times 0.500 = 0.333\)

  4. \(\theta_2 = \sin^{-1}(0.333) = 19.5°\)

The refracted angle is 19.5° (light bends toward normal).

18.4.2 Example 2: Lens equation

A converging lens has focal length 12 cm. An object is placed 30 cm from the lens.

Solution:

  1. Use the thin lens equation: \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\)

  2. Substitute: \(\frac{1}{12} = \frac{1}{30} + \frac{1}{v}\)

  3. Rearrange: \(\frac{1}{v} = \frac{1}{12} - \frac{1}{30} = \frac{5-2}{60} = \frac{3}{60} = \frac{1}{20}\)

  4. Therefore: \(v = 20\) cm

The image forms 20 cm from the lens on the opposite side (real, inverted image).

18.4.3 Example 3: Magnification

Using the lens from Example 2, find the magnification.

Solution:

  1. Use the magnification formula: \(m = -\frac{v}{u}\)

  2. Substitute: \(m = -\frac{20}{30} = -0.67\)

  3. Interpretation:

    • \(|m| < 1\): Image is smaller than object (0.67× the size)
    • \(m < 0\): Image is inverted

18.4.4 Example 4: Critical angle

Find the critical angle for light travelling from glass (\(n = 1.50\)) to air (\(n = 1.00\)).

Solution:

  1. Use the critical angle formula: \(\sin\theta_c = \frac{n_2}{n_1}\)

  2. Substitute: \(\sin\theta_c = \frac{1.00}{1.50} = 0.667\)

  3. Calculate: \(\theta_c = \sin^{-1}(0.667) = 41.8°\)

For angles greater than 41.8°, total internal reflection occurs.

18.5 Common Misconceptions

Common Misconceptions
  • Misconception: Refraction changes the frequency of light. Correction: Frequency stays constant; speed and wavelength change.

  • Misconception: A virtual image can be projected on a screen. Correction: Only real images (where rays actually converge) can be projected.

  • Misconception: All lenses magnify. Correction: Magnification can be greater than or less than 1, and images can be larger or smaller than the object.

  • Misconception: Light always bends toward the normal. Correction: Light bends toward the normal when entering a denser medium, away from the normal when entering a less dense medium.

18.6 Practice Questions

18.6.1 Easy (2 marks)

State the law of reflection.

  • Angle of incidence equals angle of reflection (1)
  • Both angles measured from the normal (1)

Answer: The angle of incidence equals the angle of reflection. Both angles are measured from the normal to the surface.

18.6.2 Medium (4 marks)

Light enters water (\(n = 1.33\)) from air at an incident angle of 40°. Find the refracted angle.

  • Apply Snell’s Law correctly (2)
  • Correct calculation: \(\sin\theta_2 = (1.00/1.33) \times \sin40° = 0.483\) (1)
  • Correct angle: \(\theta_2 = 28.9°\) (1)

Answer: 28.9° (or 29°)

18.6.3 Hard (5 marks)

A lens with \(f = 8.0\) cm forms an image 24 cm from the lens. Find the object distance and magnification.

  • Use thin lens equation: \(\frac{1}{8} = \frac{1}{u} + \frac{1}{24}\) (1)
  • Solve for \(u\): \(\frac{1}{u} = \frac{1}{8} - \frac{1}{24} = \frac{2}{24} = \frac{1}{12}\) (2)
  • Object distance: \(u = 12\) cm (1)
  • Magnification: \(m = -24/12 = -2\) (inverted, enlarged) (1)

Answer: Object distance = 12 cm; Magnification = -2 (inverted, twice as large)

18.7 Summary

Key Takeaways
  • Reflection: \(\theta_i = \theta_r\) (angles measured from normal)
  • Snell’s Law: \(n_1 \sin\theta_1 = n_2 \sin\theta_2\)
  • Thin lens equation: \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\)
  • Magnification: \(m = -v/u\)
  • Real images can be projected; virtual images cannot

18.8 Self-Assessment

Check your understanding:

After studying this section, you should be able to: